CHEMICAL ENGINEERING

HUMIDIFICATION AND WATER COOLING 321

Solution

The feed rate of wet nitrate is 1.5 kg/s containing 5.0% moisture or 1. 5 ð 5 / 100 D
0 .075 kg/s water.

∴ flow of dry solidsD 1. 5  0. 075 D 1 .425 kg/s

If the product containswkg/s water, then:

w/wC 1. 425 D 0. 2 / 100  or wD 0 .00286 kg/s

and: the water evaporatedD 0. 075  0. 00286 D 0 .07215 kg/s

The problem now consists of an enthalpy balance around the unit, and for this purpose
a datum temperature of 294 K will be chosen. It will be assumed that the flow of dry air
into the unit ismkg/s.

Considering the inlet streams:

(i) Nitrate: this enters at the datum of 294 K and hence the enthalpyD0.

(ii) Air:Gkg/s of dry air is associated with 0.007 kg moisture/kg dry air.

∴ enthalpyD[Gð 0. 99 C 0. 007 Gð 2. 01 ] 405  294 D 111. 5 GkW

and the total heat into the systemD 111. 5 GkW.

Considering the outlet streams:

(i) Nitrate: 1.425 kg/s dry nitrate contains 0.00286 kg/s water and leaves the unit at
339 K.

∴ enthalpyD[ 1. 425 ð 1. 88 C 0. 00286 ð 4. 18 ] 339  294 D 120 .7kW

(ii) Air: the air leaving contains 0.007Gkg/s water from the inlet air plus the water
evaporated. It will be assumed that evaporation takes place at 294 K.

Thus:
enthalpy of dry airDGð 0. 99  355  294 D 60. 4 GkW
enthalpy of water from inlet airD 0. 007 Gð 2. 01  355  294 D 0. 86 GkW
enthalpy in the evaporated waterD 0 .07215[2450C 2. 01  355  294 ]D 185 .6kW

and the total heat out of the system, neglecting lossesD 306. 3 C 61. 3 GkW.

Making a balance:

111. 5 GD 306. 3 C 61. 3 G or GD 6 .10 kg/s dry air

Thus, including the moisture in the inlet air, moist air fed to the dryer is:

6. 10  1 C 0. 007 D 6 .15 kg/s